ATINER CONFERENCE PRESENTATION SERIES No: EMS2017-0110 · Controllability and observability matrices. Kalman decomposition. Changes of Bases in the System Equations and Invariance

ATINER CONFERENCE PRESENTATION SERIES No: EMS2017-0110

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ATINER’s Conference Paper Proceedings Series

EMS2017-0110

Athens, 5 October 2018

Applying Dynamical Discrete Systems to Teach Mathematics

Marta Peña

Athens Institute for Education and Research

8 Valaoritou Street, Kolonaki, 10683 Athens, Greece

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ATINER’s Conference Paper Proceedings Series

EMS2017-0110

Athens, 5 October 2018

ISSN: 2529-167X

Marta Peña, Associate Professor, Polytechnic University of Catalonia, Spain

Applying Dynamical Discrete Systems to Teach Mathematics

ABSTRACT

The main goal is using examples of dynamical discrete systems in order to

illustrate basic algebraic notions such as: the matrix of a linear map, their

eigenvalues and eigenvectors. In particular: the computation of the matrix by

means of consecutive states; and the eigenvectors and eigenvalues as stationary/

asymptotic distribution and growing; equilibrium points and its stabity; stochastic

matrices. We include a guideline on discrete systems for faculty lacking in this

topic.

Keywords: dynamical discrete systems, eigenvalues, eigenvectors, equilibrium

point, stability.


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Introduction

One of the recurrent controversies at our Engineering Faculty concerns the

orientation of first-year basic courses, particularly the subject area of mathematics,

considering its role as an essential tool in technological disciplines.

The Bologna process is a good opportunity to delve into this debate. In fact,

the Faculty Council has decided that 20% of the credits from basic courses must

be related to technological applications. Thus, a mathematical engineering seminar

was held during an academic year at the Engineering Faculty of Barcelona, being

the author one of the organizers. Sessions were each devoted to one technological

discipline and aimed at identifying the most frequently used mathematical tools

with the collaboration of guest speakers from mathematics and technology

departments.

Moreover, the implementation of the Bologna process is presented as an

excellent opportunity to substitute the traditional teaching-learning model by

another one where students play a more active role. In this case, we can use the

Problem-Based Learning (PBL) method. This environment is a really useful tool

to increase student involvement as well as multidisciplinary. With PBL, before

students increase their knowledge of the topic, they are given a real situation-based

problem which will drive the learning process. Students will discover what they

need to learn in order to solve the problem, either individually or in groups, using

tools provided by the teacher or „facilitator‟, or found by themselves. In contrast

with conventional Lecture-Based Learning (LBL) methods, with the new

technique students have more responsibility since they must decide how to

approach their projects. They must identify their own learning requirements, find

resources, analyze information obtained by research, and finally construct their

own knowledge. Moreover, PBL helps students develop skills and competences

such as group and self-assessment skills, which will allow them to keep up-to-date

and continue to learn autonomously, or to become acquainted with the decision-

making process, time scheduling and, last but not least, improve communication

skills. Furthermore, theory and practice are integrated and motivation is enhanced,

which results in increased academic performance.

A collection of exercises and problems was created to illustrate the

applications identified in the seminar sessions. These exercises would be some of

the real situation-based problems given for introducing the different mathematic

topics. Two conditions were imposed: availability for first-year students and

emphasis on the use of mathematical tools in technical subjects in later academic

years. As additional material, guidelines for each technological area addressed to

faculty without an engineering background were defined. Some of them have been

already published by the author [1], [2].

The following exercises illustrate the use of Linear Algebra in different

engineering areas. As said above, they are based on conclusions drawn at the

above seminar.


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Set of complex numbers.

Alternating current representation. Its use to calculate voltage drop and

cancellation of reactive power.

Matrices. Determinants. Rank.

Controllability of Control Linear Systems. Observability of Control Linear

Systems. Its controllability indices. Composition of Systems. Realizations.

Linear System Equations.

Network flows. Leontief economic model.

Vector Spaces. Bases. Coordinates.

Color codes. Crystallography. States in Discrete Systems. Control

Functions in Discrete Systems.

Vector Subspaces.

Reachable states of Control Linear Systems. Circuit analysis (mesh

currents, node voltages).

Linear maps.

Controllability and observability matrices. Kalman decomposition.

Changes of Bases in the System Equations and Invariance of the Transfer

Function Matrix.

Diagonalization. Eigenvectors, eigenvalues.

Strain and stress tensors. Circulant matrices. Dynamical Discrete Systems.

Invariant Subspaces and Restriction to an Invariant Subspace. Controllable

Subsystem. Poles and Pole Assignment.

Non-diagonizable matrices;

Control canonical form.

Equilibrium points. Stability.

Dynamical Discrete Systems. Dynamical Continuous Systems.

Dynamical Discrete Systems.

Leslie population model. Gould accessibility indices.

Here, we present some of this material: some exercises and the guideline

for dynamical discrete linear systems. As general references regarding Linear

Algebra, see for example [3], [4].

Guidelines for Dynamical Discrete Linear Systems

We include an example of guideline for faculty lacking an applied background.

Definition of Dynamical Discrete Linear Systems

Definition

(1) A discrete linear system with constant coefficients is an equation of the

form

)()()1( kbkAxkx , 0k


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where nMA , ))(),...,(()( 1 kbkbkb n are n given discrete functions

and ),...,( 1 nxxx are n discrete functions to determine such that they verify

the above equality.

(1‟) In a more explicit form, if )( ijaA :

)(...)()1( 11111 kxakxakx nn

…

)(...)()1( 11 kxakxakx nnnnn

(2) Then it is said that x(k)is a solution with initial

conditions ))0(),...,0(()0( 1 nxxxn.

(3) If b(k) = 0,the system is called homogeneous and, otherwise it is complete.

Resolution of Dynamical Discrete Linear Systems

Proposition (Homogeneous case)

We consider a homogeneous system:

)()1( kAxkx

(1) Fixed some initial conditions )0(x , there exists a unique solution of the

system, given by:

)0()( xAkx k , 0k

(2) The set of solutions, when varying )0(x , forms a vectorial space S0 of

dimension n.

(2‟) A subset of solutions is linearly independent if and only if its initial

conditions are so.

Observation

(1) In general, the computation of the powers kA requires reducing the matrix

A to its Jordan form JA . Then

n

k

k

JJ

k

JJJJJ

c

c

J

J

SxSASkxJ

J

ASSA

1

2

1

0

1

2

1

1 )()(

We remind that

n

J

c

c

Sx 1

)0(


6

2

1

2

1

1

1

k

k

k

k

k

k

It is especially simple the diagonalible case, which we will see now. (2) If we

denote by 1)( kA ,…, n

kA )( the columns of kA , we can write the general solution as

)0()(...)0()()( 11 nn

kk xAxAkx

which shows that the columns of kA form a basis of 0S .

Corollary (Diagonalizable Homogeneous Case) – In the Above Conditions:

(1) If vx )0( is an eigenvector of A with eigenvalue , its corresponding

solution is:

vkx k)(

(2) If A diagonalizes, andv1,...,vn is a basis of eigenvectors, with respective

eigenvalues n ,...,1 , all of them real, any solution is of the form:

n

k

nn

k

n

k

n

k

n vcvc

c

c

vvkx

...)()( 111

11

1

where the coefficients ncc ,...,1 are determined by the initial conditions

)0(x :

nnvcvcx ...)0( 11

that is, they are the coordinates of )0(x in the basis of eigenvalues.

Definition

(1) The solutions of the above form

vkx k)( , k

where v is an eigenvector, of eigenvalue , are called proper modes of

the system.

(2) If 1 is a real and simple eigenvalue, and its modulus is greater than the

remainder eigenvalues

,..., 321


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it is called dominant eigenvalue and the corresponding proper mode

11)( vkx k

dominant mode. For any other solution of the form

...)( 111 vckx k , 01 c

this first addend is called its dominant part.

(2‟) Analogously, if 2 is a simple eigenvalue and

,..., 4321

it is called subdominant eigenvalue and the corresponding proper mode

subdominant mode.

Observation

The above corollary generalizes to the case of conjugate complex

eigenvalues. We suppose, for example, 12 . Then, we can take 12 vv , and

it must be 12 cc so that it has real coordinates.

Proposition (Complete Case)

Given a complete system

)()()1( kbkAxkx

the set of solutions is

00 )( SkxS

where:

- S0 is the set of solutions of the associated homogeneous

system, )()1( kAxkx .

- )(0 kx is any particular solution; for example, the one corresponding

to 0)0( x :

)1(...)1()0()( 21

0 kbbAbAkx kk

Dynamical Behaviour of the Proper Modes

First we see that the proper modes are invariant (or stationary) solutions.

Definition

(1) Given an homogeneous system

)()1( kAxkx

A subspace nF is called (dynamically) invariant if any other solution

which begins in F it is maintained inside of F:

kFkxFx ,)()0(


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It is obvious that it is equivalent to the “algebraically invariant” condition:

FFA )(

(2) The following particular cases are especially interesting:

(2.1) It is called of output (or escaping) if any other (non null) solution

in F is not bounded.

(2.2) It is called of input if any other solution in F converges to the

origin.

We analyze now that the one-dimensional invariant subspaces are spanned by

real eigenvectors of A and they correspond to real proper modes of the system. The

two-dimensional case corresponds to conjugate complex eigenvalues, as we will

see later.

Proposition (Behaviour of the Real Proper Modes)

Given an homogeneous system

)()1( kAxkx

(1) If vn is an eigenvector (and only in this case) it is verified that:

kFkxvFx ,)()0(

(1‟) In a more precise form, we remind that:

vckxcvx k )()0(

where is the eigenvalue of v.

(2) Then,

(2.1) If >1, vF is “of output”, that is:

Nkkx ),( not bounded, )0(,)0( xFx 0.

(2.2)If <1, vF is a straight line “of input”, that is:

)(kx 0, Fx )0( .

(2.3)If =1, vF is a straight line of “fixed points”, that is:

)(kx 0, Fxk )0(, .

(2.4)If =-1, all the solutions in vF are oscillating, that is:

)0()1()( xkx k , Fxk )0(, .

Observation

(1) In other words, the proper modes are the only solutions which keep the

proportion between the different coordinates )(),...,(1 kxkx n . In this sense,

they are called stationary modes. This stationary distribution is given by the

coordinates of the corresponding eigenvector.

(1‟) Then the eigenvalue is the growth rate, the same for all the coordinates,

and therefore the global for the set of population.


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(2) We will see that generically the solutions converge to the dominant mode.

So, the coordinates of the dominant eigenvector give the a symptotic

stationary distribution and the eigenvalue, the asymptotic growth rate.

As we have said, the eigenvectors of conjugate complex eigenvalues span

two-dimensional invariant subspaces, with rotatory behaviours:

Observation (Conjugate Eigenvalues: Rotatory Modes)

Given an homogeneous system

)()1( kAxkx

And we assume conjugate complex eigenvalues and their corresponding

eigenvectors: ie11 ,

ie 112

iwuv 1 , iwuvv 12

(1) The plane F is invariant:

kFkxFx ,)()0(

(1‟) More precisely. In the basis ),( wu : )0(ˆ)( 1 xekx kik

. So:

(1‟.1) If 11 , F is an escaping plane (in fact, the solutions are divergent

spirals).

(1‟.2) If 11 , F is an entry plane (in fact, the solutions are convergent

spirals to the origin).

(1‟.3) If 11 , F is a plane of turns.

(2) If 1 is simple, and ,...,, 431 the above solutions are dominant, and

the remainder ones converge asymptotically to them.

Asymptotic Convergence to the Dominant Mode

Proposition (Asymptotic Convergence to the Dominant Mode)

Given an homogeneous discrete system

)()1( kAxkx

with A diagonalizable. Being n ,...,1 the eigenvalues, and ),...,( 1 nvv a basis of

eigenvectors.

We suppose 1 is a dominant eigenvalue and )(kx is a solution, with initial

conditions

nnvcvcx ...)0( 11

Then:

(1) 111)( vckx k , for k , if 01 c .

More precisely:


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11

1

)(lim vc

kxkk

, if 01 c

(1‟) In particular, if 01 c :

1

)(

)1(lim

kx

kx

k

1

11)sgn(

)(

)(lim

v

vc

kx

kx

k

(2) Moreover, if 2 is a subdominant eigenvalue, then:

2

1

1

11

111

)1(

)(lim

vckx

vckx

k

k

k, if 02 c

Observation – We can say, then, that x(k) asymptotically converges to its

dominant part 111 vc k , with velocity of approximation 2 .

An important application of the above proposition is the following corollary,

basis of a lot of numerical algorithms for the computation of eigenvalues and

eigenvectors. The hypothesis of existence of dominant eigenvalue can be

guaranteed, for example, by means of the Perron-Frobenius theorems for matrices

with positive coefficients, as we will see. Then, moreover, the dominant

eigenvalue results positive and real (so, 11 ).

Corollary (power method for the computation of the dominant eigenvalue and

the dominant eigenvector)

We suppose that a diagonalizable matrix A has a dominant eigenvalue 1 .

Then:

1

1

lim

wA

wA

k

k

k

wA

wAk

k

klim

is a dominant eigenvector.

for wn generic (more precisely: that in a basis of eigenvectors have the first

coordinates non null).

Equilibrium Points. Stability

Finally, we study the behaviour of the solutions with regard to the equilibrium

point.

Definition – We consider a discrete system of the form

bkAxkx )()1(

(1) A constant solution cx is called an equilibrium point


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bAxx ee

(2) We assume a unique equilibrium point ex . It is called:

(2.1) Unstable if some other solution is not bounded.

(2.2) Asymptotically stable if any other solution converges to ex .

ek

xkx )(lim

(2.3) Marginally stable if any other solution is bounded, but there is some

solution not convergent to ex .

Proposition (Stability of an Equilibrium Point)

We consider a discrete system of the form

bkAxkx )()1(

(1) There exists a unique equilibrium point ex if and only if 1is not eigenvalue

of A, and then

bAIxe

1)(

(2) Then:

(2.1) If 1 for some eigenvalue of A, it is unstable.

(2.2) If 1 for all the eigenvalues of A, it is asymptotically stable.

(2.3) If 1 for all the eigenvalues of A, the system is marginally stable if

and only if the eigenvalues with 1 have the same geometric

multiplicity than algebraic multiplicity; otherwise, it is unstable.

(2‟) In particular, if there is dominant eigenvalue 1 :

(2‟.1) 11 unstable.

(2‟.2) 11 asymptotically stable.

(2‟.3) 11 marginally stable.

Observation – Analogously to (2‟) above, if 1 is a complex and simple

eigenvalue, with 12 and ,, 431 :

(1) 11 unstable.

(2) 11 asymptotically stable.

(3) 11 marginally stable.

In fact, the dominant solutions are turns around ex , and the other

generic solutions tend asymptotically to them.


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Positive Matrices

We have seen that, in the dynamical behaviour of a discrete system, the

existence of a dominant eigenvalue plays an important role. We will see that

this is guaranteed for matrices with positive coefficients:

Theorem (Perron, 1907)

We consider a discrete system

)()1( kAxkx

with A a positive matrix, that is:

0),( ijij aaA nji ,1

Then:

(1) There exists a dominant eigenvalue 1 (so, real and simple) and 01 .

(2) Its (normalized) eigenvector also has positive coordinates.

(3) It is the only (normalized) eigenvector with non-negative coordinates.

Observation

(1) The dominant eigenvalue 1 is usually called “Perron root”, and it verifies:

j

iji

j

iji

aa maxmin 1

(1‟) As a Perron eigenvector it is usually taken the one that has sum of

coordinates equal to 1 (usually called “stochastic” eigenvector).

(2) The condition 0A is frequent in population models, where the variables

must be positive. Then:

The Perron eigenvalue gives the asymptotic growth rate.

The coordinates of the stochastic eigenvalue give the asymptotic

population distribution.

(3) The above theorem is applied in particular to stochastic matrices ( the

sum of the coefficients of each columns is 1). Then the Perron eigenvalue is

1. Moreover it can be ensured that there exists k

kAA lim

and that it is a stochastic positive matrix of rank 1. In fact all the columns

are equal to the stochastic eigenvector.

(3‟) Stochastic matrices appear just when

)()(1 kxkx n constant

For example, when the global population is constant.

(4) The Frobenius theorem (1912) generalizes the Perron theorem to a wider

family of matrices 0A , the so-called “primitive” matrices.


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The Particular Case of Equations in Linear Finite Differences

Definition

(1) We will denote by y(k), k a discrete variable function (or, simply,

“discrete function”), that is, y: . It can be identified with the sequence

(y(0), y(1),…,y(k),…).

(2) An equation in linear finite differences with constant coefficients of order

n is an equation of the form:

(*))()()1(...)1()( 01 kkyakaynkyanky n

where ja are constants 10 nj , is a given discrete function and y is a

discrete function to determine such that it verifies the above relation (*).

(2‟) Then it is said that y(k) is a solution of (*), with initial conditions y(0),…,

y(n-1).

(2‟‟) If = 0, the equation in differences is called homogeneous. Otherwise it

is called complete, and the associated homogeneous equation is the one that

results of substituting )(k by 0.

(3) It is called characteristic polynomial of (*) the following one:

0111 ...)( aaaQ nn

n

Its roots are called characteristic values.

(3‟) If there is one root which is real and simple with modulus strictly greater

than the remainder ones, it is called dominant (and the following one, if there

exist, subdominant).

Observation

The equation (*) indicates the way of constructing the sequence y(k) by

recurrence, and the first objective is to find an explicit expression of the general

term y(k).

Proposition

An equation in linear finite differences

)()()1(...)1()( 01 kkyakaynkyanky n

can be considered a particular case of discrete linear system. Indeed, if we

define 1x ,…, nx by:

)1()(),...,1()(),()( 21 nkykxkykxkykx n

it results

)()()1( kbkAxkx


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)(

0

0

0

)(,

1000

0100

0010

1210 k

kb

aaaa

A

n

So, the first coordinate )(1 kx of the solution )(kx of this system is the

solution )(ky of the initial equation in linear finite differences.

We observe that A is a “companion” matrix, with characteristic polynomial

the one of the output equation in linear finite differences. So, the eigenvalues of A

are the characteristic values of the equation in linear finite differences, with the

same multiplicities.

Corollary

Given a homogeneous equation in linear finite differences:

(1) If the characteristic values are n ,...,1 , all different (which it is

equivalent to say that all the characteristic values are simple), then the general

solution is of the following form: k

nn

k ccky ...)( 11

where 1c ,…, nc are determined by the initial conditions y(0),…, y(n-1), that is:

nccy ...)0( 1

nnccy ...)1( 11

11

11 ...)1( n

nn

n ccny

More explicitly,

)1(

)1(

)0(111

11

1

12

1

ny

y

y

c

c

c

n

n

n

n

n

where the matrix is effectively invertible if n ,...,1 are different

(Vandermonde determinant).

(2) In particular, if we suppose that there exists a dominant characteristic

value 1 . Then:

1)(

)1(lim

ky

ky

k, if 01 c .

If, moreover, 2 is the subdominant characteristic value:


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2

22

11)(lim

k

k

k c

cky, if 02 c .

Exercises in Dynamical Discrete Linear Systems

We now present some exercises in dynamical discrete linear systems.

Exercise 1 (Emigration)

We consider the population exchange city center/suburbs, assuming to

simplify that the total number of habitants is constant, equal to 1.000.000. We

denote by 1x , 2x the census in the city center and the suburbs, respectively, in

thousands of habitants.

(a) If the value of 1x the years 1995, 2000 and 2005 was 600, 400 and 300,

respectively, compute the prevision for 2010, assuming that the linear

transformation which modelizes the five-year change of ),( 21 xx is the

same in all the period.

(b) Determine the matrix A of this linear map.

(c) Think on which conditions the city center will be empty and on the

contrary, when it will reach an equilibrium point. In this case, compute this

constant census and analyze if in any conditions one converges to it.

Exercise 2 (Dam/Predator)

(A) We consider the simplified model of dam/predator

kkk

kkk

PDP

PDD

1'1125'0

4'05'0

1

1

where kk DP , indicate the number of dams and predators, respectively, the year

k. We can state it in the form of a discrete system:

1'1125'0

4'05'0,)();()1( A

P

DkxkAxkx

k

k

(a) Determine the eigenvalues and eigenvectors of the matrix A .

(b) Find the solution of this model, depending on the initial conditions.

Explain the result.

(B) Now we consider the model of dam/predator depending on :


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1'1

4'05'0,)();()1(

A

P

DkxkxAkx

k

k

where represents the voracity of the predators.

(a) Prove that the eigenvalues of the matrix A are lower than 1 if and only if

1250́ .Justify that then both species converge to extinction.

(b) Justify analogously that for 1250́ the populations converge to

increase and that for 1250́ , to stabilize.

(c) Determine the distribution of the population which converges for

1250́ and 1040́ . Explain the result.

Exercise 3(American Owl)

In the study of Lamberson (1992) about the survival of the American owl,

he experimentally obtained that:

940́710́0

00180́

330́00

,

1

1

1

A

D

S

Y

A

D

S

Y

k

k

k

k

k

k

where kY , kS and kD indicates the “young” population (until 1 year old),

“subadult” population (between 1 and 2 years old) and “adult” population,

respectively, the year k.

(a) Explain the coefficients of matrix A.

(b) Verify that the eigenvalues of matrix A are approximately: 0´98,

i210́020́ .

(c) Deduce that in these conditions, the American owl converges to extinction.

(d) Verify that an improvement of the survival in the transition sub adult/adult

does not avoid extinction.

(e) In contrast, verify that extinction would be avoided if the young survival

index is 30% instead of 18%.

(Remark: the referred survival index was effectively improved by

means of appropriate forest policies.)

(f) In these conditions, compute the year increase index of the global

population, and the population distribution between young, sub adult and

adult which converge.

Exercise 4(Gould Accessibility Indices)

The “Gould accessibility indices” have been used in some geographic

problems, as for example transport nets or migration movements. For their

determination a net (or graph) is configured representing the cities (or other


17

entities geographically significant) and the connections between them. For

example:

(2)

(4) (1)

(3) Assuming that the knots or vertices are numbered (i=1,…,n), it is called its

(modified) adjacency matrix the symmetric matrix )()( RMaA nij defined

by: 1iia ; 1ija or 0, depending if the corresponding vertices are or not

connected by an arc. So, for the above graph it would be:

1001

0111

0111

1111

A

Thus, for the system )()1( kAxkx , the relation between the situations

at the instants k and k+1 is given by:

)()()()()1( 43211 kxkxkxkxkx

)()()()1( 3212 kxkxkxkx

)()()()1( 3213 kxkxkxkx

)()()1( 414 kxkxkx

More in general, we suppose that initially in each vertex there is a certain

number (not null) of objects )0(),...,0(1 nxx , and that in each unity of time each

object creates a copy in each one of the adjacent vertices, being )(),...,(1 kxkx n

the objects which are in the corresponding vertex, at the instant k. For example,

possible sequences in the above example would be:

(1,1,1,1),(4,3,3,2),(12,10,10,6),(38,32,32,18),…

(1,2,3,4),(10,6,6,5),(27,22,22,15),(86,71,71,42),…

(4,3,2,1),(10,9,9,5),(33,28,28,15),(104,89,89,48),…

It can be proved that its greater eigenvalue is positive and simple, and that we

can take as a basis of its 1-dimensional subspace associated to this eigenvalue a

vector with positive coordinates, whose sum is 1. These coordinates are called

the Gould accessibility indices of the corresponding vertices.

(a) Determine the Gould accessibility indices for the vertices of the above

graph.

(b) Analyze if the obtained result fits with the intuitive idea of “accessibility”

of each vertex.

(c) Prove that, asymptotically; the percentage of objects in each vertex is given

by the Gould accessibility indices, independently of the initial situation.


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(d) Determine, in the above sequences, the percentage of objects in each

vertex for k=3 and compare them with the obtained indices in (a).

Exercise 5 (Bus Stations)

Let us consider four bus stations A, B, C, D. The traffic is determined by the

following rules:

(i) Stations A, B: 1/3 of buses go to C; 1/3 of buses goes to D; 1/3 of buses

remains for maintenance.

(ii) Station C (respectively D): 1/4 of buses goes to A; 1/4 of buses goes to B;

1/2 of buses goes to D (respectively C).

Prove that there is asymptotic stationary distribution of the buses, and

compute it.

Exercise 6 (Engineering School)

(A) In a very demanding Engineering School there only approve the 30% of the

students each course of the degree. The remainder students repeat the course,

except in first course, where the 50% of the students leave the degree. Every year

600 new students enter in the school. Then:

(a) Determine the equations which govern the number of students every

course.

(b) Compute the equilibrium point and study its stability.

(B) Considering the excessive massification, a very drastic change in the

pedagogical system inverted the percentages, in such a way that the percentage of

students that approve every course was 70%. These better expectations reduce the

abandonments at first course up to 10%. Then:

(a) Determine the new equations which govern the number of students

every course.

(b) Compute the new equilibrium point and study its stability.

Exercise 7 (Fibonacci Numbers)

The Fibonacci numbers are:

1,1,2,3,5,8,13,21,…

It is probably the first recurrent equation known in history which appears at

the “Liber Abaci” from 1202, collecting a problem of rabbit breeding, stated and

solved by Leonardo of Pisa.


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These numbers are given by the Fibonacci equation, which is an equation in

linear finite differences:

)()1()2( kykyky

with initial conditions:

1)1(,1)0( yy

Then:

(a) Determine the solution of this equation.

(b) Compute )(

)1(lim

ky

ky

k

.

Solution of Exercises

Solution of Exercise 1 (Emigration)

(a) Being

700

300)2(,

600

400)1(,

400

600)0( xxx

Then:

)1(2

3)0(

2

1)2( xxx

Hence

750

250)2(

2

3)1(

2

1)3( xxx

(b) Acting as in (a) for the vectors

0

1 and

1

0, we obtain the columns of:

90́40́

1'06'0A ,

which is a stochastic matrix.

(c)

e

e

e

e

x

x

x

x

2

1

2

1

90́40́

1'06'0if and only if 2001 ex .

It corresponds to the Perron eigenvalue 1, so that it is the asymptotic

distribution.

Solution of Exercise 2 (Dam/Predator)

(A)

(a) The eigenvalues and eigenvectors of A are


20

11

5

41v

6'02

1

42v

The first one indicates a stationary distribution of 4 predators for each 5

dams, which maintains the total population constant )1( 1 .

The second one indicates another stationary distribution (4 predators for

each dam), with a yearly decrease of the total population of 40% )6'0( 2 . It

fits with the appreciation that an excess of predators provoques a decrease of

dams, and as a consequence of predators as well.

We will see now that, being 11 dominant, we will converge generically

to the first situation.

(b) The solutions are of the form

222111)( vcvckx kk

with 11 dominant eigenvalue. The convergence to the dominant

mode, if 01 c , is clear so that:

02

122222

vcvc

k

k

The condition 01 c depends on the initial conditions:

2211)0( vcvcx

(b.1) 01 c only when the initial population distribution is 4/1 of 2v .

Then )(kx mantains this distribution, with a yearly decrease of 40%

for both species, converging to extinction.

(b.2) For any other initial distribution is 01 c , and then )(kx converges to

the dominant mode. But we have to distinguish the cases 01 c and

01 c , which correspond to the initial proportion of predators being

lower and greater, respectively, to the above 4/1:

0444205

44

)0(

)0(

1

412121

21

21

2

1

ccccc

cc

cc

x

x

When the initial proportion of predators is lower tan 4/1, the solution

converges to the stabilization of the total population, with an

asymptotic population distribution 4/5.

When the initial proportion of predators is greater than 4/1,

then 01 c and the solution converges to negative coordinates, which

means the extinction of both species in short term. For example,

for )1,20()0( x results 11 c , 62 c and )3(x has negative

coordinates.


21

(B)

(a) Now the dominant eigenvalue is

4'09'08'01

The dominant mode implies the stabilization of the populations when

11 , which corresponds to the value 125'0 above.

For greater voracities, the dominant eigenvalue is lower than 1, and the

populations converge to extinction for any initial conditions.

(b) For lower voracities, the dominant mode supposes the increase of the

populations, with a bigger proportion of dams than in the case

125'0 .

(c) For example, for 104'0 is 02'11 and )13,10(1 v : the

populations yearly increase 2%, and the proportion dams/predators is

1‟3 (instead of 1‟25 when 125'0 ).

Solution of Exercise 3 (American Owl)

(a) The first row is formed by the birth rate. So, the young and sub adult

population does not procreate, while each adult couple has, on average,

2 children each 3 years.

The coefficients 0´18 and 0´81 are the survival indices of the transition

young/sub adult and sub adult/adult, respectively. It is clearly

confirmed that the first one is critical: when the young phase finishes,

they have to leave the nest, find a hunting domain, find a couple,

construct a nest, etc.

The coefficient 0´94 indicates that the adult population has a yearly

death rate of 6%.

(b) Direct computation. In particular we observe that trA and detA are,

respectively, the sum and the product of the stated eigenvalues.

(c) We have seen that any solution converges to the origin when 1 for

all eigenvalues.

(d) The extinction is avoided if and only the dominant eigenvalue is greater

than 1: 1DOM

We have seen that:

980́710́32 DOMa

If we increase this survival index, the dominant eigenvalue will increase,

but it will not become 1:

99950́001́

99630́940́

99140́850́

32

32

32

DOM

DOM

DOM

a

a

a

(e) For 300́21 a , instead of 0´18, the eigenvalues become 1´01,

i260́030́ . So:

1011́ DOM


22

(f) The asymptotic increase is given by the dominant eigenvalue, so it

would yearly converge to 1%, for each cohort of population.

The asymptotic population distribution between the 3 cohorts is given

by the coordinates of the eigenvector corresponding to the dominant

eigenvalue:

)21,3,10(DOMv .

That is, for each 10 young owls, there will be 3 sub adult owls and 31 adult

owls, with a growth rate of 1%.

Solution of Exercise 4 (Gould Accessibility Indices)

(a) The eigenvalues and eigenvectors of matrix A are:

)280́,520́,520́,610́(173́ 11 v

)820́,370́,370́,250́(311́ 22 v

)0,710́,710́,0(0 33 v

)510́,300́,300́,750́(480́ 44 v

The Gould accessibility indices would be given by the coordinates of

the dominant eigenvalue, 1v , normalized to sum 1:

)14´0,27´0,27´0,320́(1 v

(b) It matches that the vertex 1 is the best connected and the 4, the worst.

Equally that the 2 and 3 have the same index.

It is less intuitive that the index of 2 and 3 is almost the double than the

one of 4, and only a 20% below than the one of 1.

(c) Asymptotically, the objects converge to be distributed according to the

coordinates of the dominant eigenvalue, 1v .

(d) If we normalize (to sum 1) x(3) for the different initial conditions, we

obtain:

)140́,270́,270́,310́()3(),1,2,3,4()0(

)16´0,26´0,260́,320́()3(),4,3,2,1()0(

)150́,270́,270́,31´0()3(),1,1,1,1()0(

xx

xx

xx

which is very similar to the Gould indices 1v .

Solution of Exercise 5 (Bus Stations)

Clearly the total number of buses is constant, so that the matrix of the discrete

system will be stochastic. Indeed, the number of buses )(kxA , )(kxB , )(kxC and

)(kxD in the respective station at the day k, is determined by


23

)(

)(

)(

)(

02/13/13/1

2/103/13/1

4/14/13/10

4/14/103/1

)1(

)1(

)1(

)1(

kx

kx

kx

kx

kx

kx

kx

kx

D

C

B

A

D

C

B

A

The a symptotic stationary distribution will correspond to the eigenvector

of the eigenvalue 11 DOM :

4

4

3

3

1 DOMvv

More explicitly, if N is the total number of buses, the asymptotic stationary

distribution is

4

4

3

3

14

N

Additionally we can check that for the other eigenvalues: 1 and each

eigenvector has negative and positive coordinates.

3/12

0

0

1

1

2v

2/13

1

1

0

0

3v

6/14

1

1

1

1

4v

Solution of Exercise 6 (Engineering School)

(A)

(a) The equations that govern the number of students )(kxi , 41 i , at

the course i and the year k are:


24

600)(20́)1( 11 kxkx

)(70́)(3́0 )1( 212 kxkxkx

)(70́)(3́0 )1( 323 kxkxkx

)(70́)(3́0 )1( 434 kxkxkx

that is,

bkAxkx )()1(

where

0

0

0

600

,

70́30́00

070́30́0

0070́30́

00020́

,

)(

)(

)(

)(

)(

4

3

2

1

bA

kx

kx

kx

kx

kx

(b) As 1 is not an eigenvalue of A, there exists an unique equilibrium point

bAIxe

1)( . In this case it is easy to compute it directly:

434

323

212

11

)(70́)(30́)(

)(70́)(30́)(

)(70́)(30́)(

600)(20́)(

eee

eee

eee

ee

xxx

xxx

xxx

xx

which has as unique solution

4,...,1,750)( ix ie

So, in the stationary state there will be 750 students in every course, with a

total number of students of 3000.

As the eigenvalues of A are 0´7, triple eigenvalue, and 0´2, all of them are

smaller than 1, the equilibrium point is asymptotically stable, so that the

solutions converge to it for any initial conditions. For example:

,...749)4(,744)3(,720)2(,600)1(0)0( 1111 xxxxx

(B)

(a) The new system is:

bkxAkx )(´)1(

where

0

0

0

600

,

30́70́00

030́70́0

0030́70́

00020́

´,

)(

)(

)(

)(

)(

4

3

2

1

bA

kx

kx

kx

kx

kx

(b) The new equilibrium point is the same than above, and it is also

asymptotically stable. But now the yearly number of degree students is 525,

instead of 225 above.


25

Solution of Exercise 7 (Fibonacci Numbers)

(a) It corresponds to a homogeneous equation in linear finite differences.

Its characteristic polynomial is:

1)( 2 Q

so that the characteristic values are:

2

51,

2

5121

and the general solution is: kk

ccky

2

51

2

51)( 21

where

2

51

2

511

0

21

21

cc

cc

so that

5

1,

5

121

cc

Hence,

,...2,1,2

51

2

51

5

1)(

kky

kk

Surprisingly, this expression gives natural number for all k :

1,1,2,3,5,…

Alternatively, it can be seen as the homogeneous discrete linear system

)()1( kAxkx

where:

11

10A

The eigenvalues of A are, equivalently:

2

51,

2

5121

so that it diagonalizes:

,0

01

2

1ASSA

where

21

11

S

and it is verified that:


26

1

2

1

0

0

SSA

k

k

k

Actually, we are interested in the value of )(1 kx for the initial conditions

1)1(,1)0( 21 xx , so that:

1

1

0

0

)(

)(1

2

1

2

1SS

kx

kxk

k

which gives:

kkkx 2115

1)(

So it coincides with the solution obtained before.

(b) It results

2

51

)(

)1(lim

ky

ky

k,

which is the famous “golden ratio”. It is basic in the Greek aesthetics,

and very frequent in natural phenomena as spirals in snails, plant

branching, etc.

Conclusions

This paper confirms the possibility of illustrating in Engineering degrees

concepts and basic results of Linear Algebra, such as: the computation of the

matrix by means of consecutive states; and the eigenvectors and eigenvalues as

stationary/asymptotic distribution and growing; equilibrium points and its stabity,

stochastic matrices, etc. In particular, we present examples of dynamical discrete

systems, which are motivating application exercises of technological disciplines.

The exercises are accessible to early-year students since they are self-contained in

terms of technological requirements and only basic knowledge of Linear Algebra

is necessary. Furthermore, they can be implemented by means of PBL

methodology.

References

[1] Ferrer, J., Ortiz, C., Peña, M. Learning Engineering to Teach Mathematics, 40

th

ASEE/IEEE Frontiers in Education Conference, 2010.

[2] Ferrer, J., Ortiz, C., Peña, M. Learning Automation to Teach Mathematics,

Automation, ISBN 978-953-51-0685-2, 2012.

[3] Lay, D. C. Algebra lineal y sus aplicaciones [Linear algebra and its applications.]

México, Ed. Pearson, 2007.


27

[4] Puerta Sales, F. Álgebra Lineal [Linear Algebra.] Universidad Politécnica de

Barcelona ETS Ingenieros Indutriales de Barcelona con la colaboración de

MARCOMBO de Boixareu Ed., 1976.

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